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An analytical chemist is titrating 99.2 mL of a 0.6300 M solution of dimethylamine ((CH3 NH) with a 0.2500 M solution of HNO3. The p K, of dimethylamine is 3.27. Calculate the pH of the base solution after the chemist has added 213.6 mL of the HNO3 solution to it.

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Answer:

pH of the base solution after the chemist has added 213.6 mL of the HNO3 solution to it is 9.96

Explanation:

Calculating the moles of dimethylamine  

Moles of dimethylamine  = molairty * volume in L = 0.63*0.0992   = 0.0625moles

Moles of HNO3               = molarity * volume in L = 0.25*0.2136    = 0.0534 moles

The balanced chemical equation is

(CH3)2NH + HNO3 -------------------> (CH3)2NH2^+    +   NO3^-

I                  0.0625         0.0534                               0                  

C                 -0.0534       - 0.0534                              0.0534

E               0.0091               0                                   0.0534

As we know ,  

POH   = Pkb + log[(CH3)2NH2^+]/[(CH3)2NH]

= 3.27 + log0.0534/0.0091                                  

= 3.27+ 0.768

= 4.038                  

PH      = 14-POH

= 14-4.038   = 9.96  

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